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 Mobile Phone Muse (Posted on 2014-05-05)
Five friends Andy, Brad, Cole, Dan and Enoch went to a supermarket and bought five sets of mobile phones.
• Each of the five friends took precisely one set. However, they chose to use mobile service networks of four service providers P, Q, R or S.
• Mobile phone sets of each person had a unique facility. These facilities were: 4k colour display, Group SMSing, Graphics and Screen savers.
It is known that:
1. Amongst, Andy, Brad, Cole, Dan and Enoch two were enjoying services of network P and two were enjoying 4k colour display facilities.
2. The one who was enjoying "group SMSing" facility in his mobile phone set, was also enjoying the Q network.
3. Neither of the two who were enjoying network P were Andy or Cole. Nor were they enjoying "4k colour display" facility in their mobile phone sets.
4. Brad who was not using the network S, was enjoying "4k colour display" facility in his mobile phone set.
5. Neither Andy nor Enoch was enjoying "4k colour display" facility or “graphics” facility.
From the above clues, determine which friend used which network and what was the facility provided in his mobile phone set.

 No Solution Yet Submitted by K Sengupta No Rating

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 computer solutions Comment 2 of 2 |
In the following seven sets of solutions, each line is in the order Andy, Brad, Cole, Dan and Enoch. The first row indicates the carrier, and the second row the feature, where 1 is 4k color display; 2 is Group SMS, 3 is graphics and 4 is screen saver.

`qprps21314qprsp21314qpspr21314qpsrp21314qrspp21314rpspq41312sprpq41312`

DefDbl A-Z
Dim used(4), netw As String, had2

Text1.Text = ""
crlf\$ = Chr(13) + Chr(10)
netw = "ppqrs": h\$ = netw
f\$ = Space\$(5)
Do
For feata = 1 To 4
used(feata) = 1
Mid(f\$, 1, 1) = LTrim(Str(feata))
For featb = 1 To 4
If used(featb) <= 2 And (used(featb) <= 1 Or had2 = 0) Then
used(featb) = used(featb) + 1
If used(featb) > 1 Then had2 = 1
Mid(f\$, 2, 1) = LTrim(Str(featb))
For featc = 1 To 4
If used(featc) <= 2 And (used(featc) <= 1 Or had2 = 0) Then
used(featc) = used(featc) + 1
If used(featc) > 1 Then had2 = 1
Mid(f\$, 3, 1) = LTrim(Str(featc))
For featd = 1 To 4
If used(featd) <= 2 And (used(featd) <= 1 Or had2 = 0) Then
used(featd) = used(featd) + 1
If used(featd) > 1 Then had2 = 1
Mid(f\$, 4, 1) = LTrim(Str(featd))
For feate = 1 To 4
If used(feate) <= 2 And (used(feate) <= 1 Or had2 = 0) Then
used(feate) = used(feate) + 1
If used(feate) > 1 Then had2 = 1
Mid(f\$, 5, 1) = LTrim(Str(feate))

good = 1
If used(1) < 2 Then good = 0
If used(2) > 1 Then good = 0
If used(3) > 1 Then good = 0
If used(4) > 1 Then good = 0
ix = InStr(netw, "q")
If Mid(f\$, ix, 1) <> "2" Then good = 0
If Mid(netw, 1, 1) = "p" Or Mid(netw, 3, 1) = "p" Then good = 0
If Mid(f\$, 1, 1) = "1" Or Mid(f\$, 3, 1) = "1" Then good = 0
If Mid(netw, 2, 1) = "s" Or Mid(f\$, 2, 1) <> "1" Then good = 0
If Mid(f\$, 1, 1) = "1" Or Mid(f\$, 5, 1) = "1" Then good = 0
If Mid(f\$, 1, 1) = "3" Or Mid(f\$, 5, 1) = "3" Then good = 0
If good Then
Text1.Text = Text1.Text & netw & crlf\$
Text1.Text = Text1.Text & f\$ & crlf\$ & crlf\$
End If

used(feate) = used(feate) - 1
If used(feate) = 1 Then had2 = 0
End If
Next
used(featd) = used(featd) - 1
If used(featd) = 1 Then had2 = 0
End If
Next
used(featc) = used(featc) - 1
If used(featc) = 1 Then had2 = 0
End If
Next
used(featb) = used(featb) - 1
If used(featb) = 1 Then had2 = 0
End If
Next
used(feata) = 0
Next

permute netw
Loop Until netw = h\$
End Sub

 Posted by Charlie on 2014-05-05 13:48:11

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